Alnico is a strongly magnetic alloy (solid mixture) of Co, Ni, Al and Fe. A 100.0 g sample of Alnico has 11.2g Al, 22.3g Ni, and 5.0g Co in it; the rest is Fe. How many moles of aluminum are there?

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Alnico is a strongly magnetic alloy (solid mixture) of Co, Ni, Al and Fe. A 100.0 g sample of Alnico has 11.2g Al, 22.3g Ni, and 5.0g Co in it; the rest is Fe. How many moles of aluminum are there?

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Answer 1 · 2016-02-29T10:44:12

$0.423$ $0.423$ $m o l$ $mol$ Al give or take w/ sig figs

Explanation:

$11.2$ $11.2$ $g$ $g$ Al x ( $1$ $1$ $m o l$ $mol$ / $26.98$ $26.98$ $g$ $g$ )= $0.415$ $0.415$ $m o l$ $mol$ Al

$22.3$ $22.3$ $g$ $g$ Ni x( $1$ $1$ $m o l$ $mol$ / $58.69$ $58.69$ $g$ $g$ )= $0.380$ $0.380$ $m o l$ $mol$ Ni

$5.0$ $5.0$ $g$ $g$ Co x ( $1$ $1$ $m o l$ $mol$ / $58.93$ $58.93$ $g$ $g$ )= $0.0848$ $0.0848$ $m o l$ $mol$ Co

$100$ $100$ $g - 11.2$ $g -11.2$ $g$ $g$ Al $- 22.3 g$ $-22.3g$ $N i - 5.0$ $Ni -5.0$ $g$ $g$ Co = $61.5$ $61.5$ $g$ $g$ Fe

$61.5$ $61.5$ $g$ $g$ Fe $\times (\frac{1 m o l}{55.84}) = 1.10$ $xx((1 mol)/55.84)=1.10$ $m o l$ $mol$ Fe

Divide each one by the smallest number of moles: cobalt at $0.0848$ $0.0848$ $m o l$ $mol$ .

Al = $\frac{0.415}{0.0848} = 4.89$ $0.415/0.0848 = 4.89$ $m o l$ $mol$
Ni = $\frac{0.380}{0.0848} = 4.48$ $0.380/0.0848 = 4.48$ $m o l$ $mol$
Co = $\frac{0.0848}{0.0848} = 1.00$ $0.0848/0.0848 = 1.00$ $m o l$ $mol$
Fe= $\frac{1.10}{0.0848} = 12.97$ $1.10/0.0848= 12.97$ $m o l$ $mol$

Since Ni is $4.48 \approx 4.50$ $4.48 ~~4.50$ , we multiply everything by $2$ $2$ to get a whole number, $9$ $9$ , for Ni.

Al = $4.89$ $4.89$ $m o l \times 2 = 10$ $mol xx 2= 10$ $m o l$ $mol$ Al
Ni = $4.48$ $4.48$ $m o l \times 2 = 9$ $mol xx 2= 9$ $m o l$ $mol$ Ni
Co = $1.00$ $1.00$ $m o l \times 2 = 2$ $mol xx 2= 2$ $m o l$ $mol$ Co
Fe= $12.97$ $12.97$ $m o l \times 2 = 26$ $mol xx 2= 26$ $m o l$ $mol$ Fe

This gives us an empirical formula of $A l_{10} C o_{2} F e_{26} N i_{9}$ $Al_10Co_2Fe_26Ni_9$ .

If we add up the Fe, Co, Ni, and Al in the empirical formula, we get $2366.7$ $2366.7$ $g m o l^{- 1}$ $gmol^-1$ of Alnico

$100$ $100$ $g$ $g$ Alnico x ( $1$ $1$ $m o l$ $mol$ Alnico/ $2366.7$ $2366.7$ $g$ $g$ Alnico) x ( $10$ $10$ $m o l$ $mol$ Al/ $1$ $1$ $m o l$ $mol$ Alnico) = $0.423$ $0.423$ $m o l$ $mol$ Al

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Alnico is a strongly magnetic alloy (solid mixture) of Co, Ni, Al and Fe. A 100.0 g sample of Alnico has 11.2g Al, 22.3g Ni, and 5.0g Co in it; the rest is Fe. How many moles of aluminum are there?

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